Flow rate across a fluid element with top and bottom faces moving

Fluid elementdy)dx.

x

qq( xx

dyqx

dx)dy.y

qq( y

y

dxqy

dxdyw 2

dxdyw1

h

dx

dy

•Consider the fluid element with dimensions dx, dy, and h (film thickness)

•Consider bottom face moving up with velocity w2 and top face moving up with velocity w1

Rate of flow out in x direction

Rate of flow in, in x direction

Rate of flow out in y direction

Rate of flow in, in y direction

Volume change in z direction due to w2

Volume change in z direction due to w1

Top and bottom faces moving vertically

dy)dx.x

qq(dyq xxx

dx)dy.y

qq(dxq y

yy

dxdyww 21

Net rate inflow in x direction =

Net rate inflow in y direction =

Rate volume change in z direction =

Therefore equating the net rate inflow in x and y directions to the rate of volume change in z direction we get

)7...(012

wwy

q

x

q yx

Top and bottom faces moving vertically

)3.....(2

h)UU(

x

p.

12

hudzq 21

3h

0

x

)4.....(2

h)VV(

y

p.

12

hvdzq 21

3h

0

y

We have seen earlier that

Substituting into eqn. (7) we get:

0)(.122

)(.122

)( 12

3

21

3

21

wwy

phhVV

yx

phhUU

x

Top and bottom faces moving vertically

)8...()(2)()(6

.12

.12

122121

33

wwdy

dhVV

x

hUU

y

ph

yx

ph

x

Assuming viscosity is constant, U1 and U2 do not vary with x, and V1 and V2 do not vary with y

Which on rearranging and simplification gives:

Net vertical velocity of top surface

dx

dhtan

y

hV

x

hUww 11h1

wh

U1

dx

dh

Downward velocity due to U1 is U1tan(new position of surface is lower than original position)

Vertical externally applied velocity = wh

Therefore net vertical velocity of upper surface =

wh – U1tan = wh – U1dh/dx

If V1 and are both non zero, then the net vertical velocity of the top

surface will be:y

h

Both surfaces inclined and translating

oo dx

dhtan

wh

U1

dx

dh

wo

U2

Bottom surface: Horizontal velocity U2, angle of inclination o, vertically applied velocity wo

Subscript “o” refers to the bottom surface

Therefore, similar to the top surface, net velocity of the bottom surface in the vertical direction:

o

2o

2o2 y

hV

x

hUww

Both surfaces inclined and translating

Substituting for w1 and w2 in Reynold’s equation which is

We get

)ww(2dy

dh)VV(

x

h)UU(6

y

p.

12

h

yx

p.

12

h

x 212121

33

)9...(

y

hV

x

hUw

y

hV

x

hUw2

dy

dh)VV(

x

h)UU(6

y

p.

12

h

yx

p.

12

h

x

o

2o

2o11h2121

33

8

P1. Velocity at a given co-ordinate

2212 U

h

z)UU()zhz(

x2

pu

Bottom surface moves with velocity U

Upper surface is stationary

h

Calculate the velocity of flow at a height 0.5 mm above the bottom surface when distance of separation is 0.8 mm. The variation of pressure with x at distance of separation 0.8 mm was found experimentally as 10 Pa/cm.

U = 5 cm/s, = 50 cp.

NOTE: 1 cp = 0.001 Pa.s

1 Pa = 1N/m2

z

h

z=0

z=hU1

U2

9

Solution P1

s/m10x15.0]10x)8.0x5.05.0[(x10x50x2

10x10)zhz(

x2

p 3623

22

s/cm875.110x5x)625.01()10x5(8.0

5.0x)10x5(U

h

z)UU( 222

221

Therefore velocity u = 1.875-0.015 = 1.86 cm/s

2212 U

h

z)UU()zhz(

x2

pu

10

P2. Finding volume flow rate

Top surface moves

h

The top surface is part of a rotating journal rotating at 120 RPM. The diameter of the journal is 20 cm. The angle of inclination of the wedge formed is = 15o The minimum height of separation is 10 m. Using Reynold’s equation in 2 dimensions, the pressure variation in the x direction at a distance x = 20 m is measured as 10 Pa/cm. Calculate the volume flow rate. = 50 cp.

(consider velocity of top surface = speed of journal circumference)

NOTE: 1 cp = 0.001 Pa.s

1 Pa = 1N/m2

z h

z=0

z=hU1

U2x

)3.....(2

h)UU(

x

p.

12

hudzq 21

3h

0

x

11

Solution problem 2

Velocity of top surface U = 2r . RPM/60 = 2r.120/60 = 1.248 m/sx = 20.10-6 m, h = xtan15o + 10.10-6 = 15.36.10-6 m

Therefore

)3.....(2

h)UU(

x

p.

12

hudzq 21

3h

0

x

23

366

x 10x10.10x50x12

)10x36.15(

2

10x36.15.248.1q

scm /106.9 36